Exercise 18, § 5

Let $k$ be a field, $A$ a finite type $k$-algebra. Now prove that $\mathrm{Reg}(A)$ is an open subset of $\mathrm{Spec}(A)$. (Using Exercise 16, reduce to proving that $\mathrm{Reg}(A)$ contains a nonempty open subset when $A$ is integral. Using the normalization lemma (V, § 3, No. 1, Theorem 1) and Exercise 17, a), reduce to proving this last property when $A$ is the integral closure of $k[X_1, ..., X_n]$ in a finite quasi-Galois extension $K$ of $k(X_1, ..., X_n)$. Let $K'$ be the radicial closure of $k(X_1, ..., X_n)$ in $K$ and $A'$ the integral closure of $k[X_1, ..., X_n]$ in $K'$. Reduce to proving the required property for $A'$ thanks to Exercise 17, b). Following the proof of Theorem 2 of V, § 3, No. 2, include $K'$ in an extension $k'(X_1^{q^{-1}}, ..., X_n^{q^{-1}}) = K''$ where $k'$ is a radicial extension of $k$ and reduce to the case where $A'$ is the integral closure of $k[X_1, ..., X_n]$ in $K''$ thanks to Exercise 17, a). Notice then that $A' = k'[X_1^{q^{-1}}, ..., X_n^{q^{-1}}]$ (V, § 1, No. 3, Corollary 2 to Proposition 13) and that $A'$ is regular (p. 94, Exercise 6).)